7.4 Mixture & Weighted Average Word Problems

Concentration and Conserved Solute

Mixture word problems sit at the overlap of OG Math Review 3.3 (rates, ratios, percents) and 3.4 (statistics: weighted averages). A solution is solute plus solvent. Concentration is the ratio

C = (amount of solute) / (total volume), often written as a percent.

Solute amount is then C × V. When two solutions mix and nothing reacts or evaporates, solute is conserved:

C₁V₁ + C₂V₂ = C_mix (V₁ + V₂)

so the mixture concentration is the volume-weighted average

C_mix = (C₁V₁ + C₂V₂) / (V₁ + V₂).

C_mix always lies strictly between C₁ and C₂. If you mix 10% acid with 40% acid, the result cannot be 5% or 45%. Cross those choices off before you algebra. The same bound is the average-speed lesson in different clothes: a weighted average cannot fall outside the data.

Worked example. Mix 20 liters of 10% acid with 30 liters of 40% acid. Solute: 0.10×20 + 0.40×30 = 2 + 12 = 14 liters of acid. Total volume 50 liters. C_mix = 14/50 = 0.28 = 28%. Check: 28 sits between 10 and 40, closer to 40 because more of the 40% solution was used.

Units: keep C as a decimal (0.20) or as a percent (20) consistently. Mixing 20 with 0.50 in the same equation is a scaling error of 100. On a no-calculator exam, clearing percents by using 20 and 50 (and 30 as the target) is often cleaner than introducing 0.20.

Alloys are the same model with metals as solute. A 12-gram alloy that is 40% copper contains 4.8 grams of copper. Mixing alloys conserves each metal separately; write one conservation equation per metal you care about.

Alligation, the Seesaw Ratio

When the question asks how much of each solution to hit a target concentration, alligation (the seesaw, or allegation) avoids solving for x in a linear equation — though the equation is still the justification.

From C₁V₁ + C₂V₂ = C_mix(V₁ + V₂), rearrange:

V₁ / V₂ = |C₂ − C_mix| / |C₁ − C_mix|.

The volume of each component is proportional to the gap from the other component to the target. Graphically, write C₁ and C₂ on the left, C_mix in the middle, and subtract diagonally:

• Parts of the C₁ solution = |C₂ − C_mix|
• Parts of the C₂ solution = |C₁ − C_mix|

Worked example. Make 60 liters of 30% acid from a 20% solution and a 50% solution. Diagonal gaps: |50 − 30| = 20 parts of 20%, and |20 − 30| = 10 parts of 50%. Ratio 20 : 10 = 2 : 1. Total 3 parts = 60 liters, so 1 part = 20 liters. Use 40 liters of 20% and 20 liters of 50%. Conservation check: 0.20×40 + 0.50×20 = 8 + 10 = 18, and 18/60 = 0.30.

The weaker solution takes more volume when the target is closer to the weaker concentration — 30 is closer to 20 than to 50, so the 2 : 1 ratio favors the 20% stock. That is the same "closer to the slower speed" intuition from the harmonic mean, now in concentration units.

Alligation also handles mixing a solution with water (C₂ = 0) or with pure solute (C₂ = 100). You do not need a separate theory; 0 and 100 are legal concentrations on the seesaw.

Dilution, Fortification, and Evaporation

Dilution. C₁V₁ + 0 = C_final(V₁ + w). The same solute sits in a larger tank, so concentration falls. Worked example: 40 ounces of 15% alcohol contains 6 ounces of alcohol. Adding w ounces of water to reach 10%: 6 / (40 + w) = 0.10, so 6 = 4 + 0.10w and w = 20 ounces. Check: 6/60 = 10%.

Fortification. C₁V₁ + 1·x = C_final(V₁ + x). Worked example: 40 liters of 10% salt, add x liters of pure salt to reach 20%. Then (4 + x)/(40 + x) = 1/5, so 5(4 + x) = 40 + x, 20 + 5x = 40 + x, 4x = 20, x = 5 liters. Alligation: 10% and 100% around a 20% target. Gaps: |100 − 20| = 80 parts of 10%, |10 − 20| = 10 parts of 100%, ratio 8 : 1. With 40 liters of 10%, pure salt is 40/8 = 5 liters.

Evaporation. Water leaves; solute stays. C₁V_initial = C_final V_final, so V_final = (C₁ / C_final) × V_initial, and water boiled off is V_initial − V_final. Worked example: 40 ounces at 15% alcohol, evaporate to 25%. Solute still 6 ounces, so 6 = 0.25 V_final and V_final = 24 ounces. Water removed: 40 − 24 = 16 ounces. Evaporation raises concentration; dilution lowers it. If a choice moves the wrong way, it is impossible.

A replacement that removes mixture (not pure water) and then adds water is not evaporation. That process removes some solute. Use the next formula.

Repeated Removal and Replacement

A container of volume V holds Q₀ of solute (often it starts full of pure solute, so Q₀ = V). You remove x units of well-mixed liquid and replace those x units with pure water, n times.

Each cycle leaves the fraction (1 − x/V) of whatever solute was present:

Q_n = Q₀ (1 − x/V)^n

and C_n = C₀ (1 − x/V)^n because the volume returns to V after every replacement.

Worked example. A 60-liter tank of pure alcohol. Remove 12 liters, replace with water, three times. Multiplier per cycle: 1 − 12/60 = 4/5. After three cycles: 60 × (4/5)³ = 60 × 64/125 = 30.72 liters of alcohol. Cycle-by-cycle: 48, then 38.4, then 30.72. Stopping after two cycles and reading 38.4 is the usual partial-count trap.

Expand small n with fractions rather than decimals when the exam's choices are exact. (3/4)² = 9/16 is safer than 0.75² if you are already carrying 80 liters: 80 × 9/16 = 45 exactly.

Do not treat repeated replacement as n times a linear subtraction of x units of pure solute. After cycle 1 the liquid you remove is already diluted, so each later removal takes less solute than the one before. That is why the factor is a product of (1 − x/V), not a sum.

Setup sheet